3.492 \(\int \frac {\tanh (e+f x)}{(a+b \sinh ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=69 \[ \frac {1}{f (a-b) \sqrt {a+b \sinh ^2(e+f x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a-b}}\right )}{f (a-b)^{3/2}} \]

[Out]

-arctanh((a+b*sinh(f*x+e)^2)^(1/2)/(a-b)^(1/2))/(a-b)^(3/2)/f+1/(a-b)/f/(a+b*sinh(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3194, 51, 63, 208} \[ \frac {1}{f (a-b) \sqrt {a+b \sinh ^2(e+f x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a-b}}\right )}{f (a-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[e + f*x]/(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

-(ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]]/((a - b)^(3/2)*f)) + 1/((a - b)*f*Sqrt[a + b*Sinh[e + f*x]^
2])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\tanh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{(1+x) (a+b x)^{3/2}} \, dx,x,\sinh ^2(e+f x)\right )}{2 f}\\ &=\frac {1}{(a-b) f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{2 (a-b) f}\\ &=\frac {1}{(a-b) f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^2(e+f x)}\right )}{(a-b) b f}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{3/2} f}+\frac {1}{(a-b) f \sqrt {a+b \sinh ^2(e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 58, normalized size = 0.84 \[ -\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b \cosh ^2(e+f x)}{a-b}+1\right )}{f (b-a) \sqrt {a+b \cosh ^2(e+f x)-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[e + f*x]/(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

-(Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*Cosh[e + f*x]^2)/(a - b)]/((-a + b)*f*Sqrt[a - b + b*Cosh[e + f*x]^2]
))

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fricas [B]  time = 0.66, size = 1370, normalized size = 19.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/2*((b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(2*a - b)*cosh(f*x + e)^
2 + 2*(3*b*cosh(f*x + e)^2 + 2*a - b)*sinh(f*x + e)^2 + 4*(b*cosh(f*x + e)^3 + (2*a - b)*cosh(f*x + e))*sinh(f
*x + e) + b)*sqrt(a - b)*log((b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(4
*a - 3*b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 4*a - 3*b)*sinh(f*x + e)^2 + 4*sqrt(2)*sqrt(a - b)*sqrt((
b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x +
e)^2))*(cosh(f*x + e) + sinh(f*x + e)) + 4*(b*cosh(f*x + e)^3 + (4*a - 3*b)*cosh(f*x + e))*sinh(f*x + e) + b)
/(cosh(f*x + e)^4 + 4*cosh(f*x + e)*sinh(f*x + e)^3 + sinh(f*x + e)^4 + 2*(3*cosh(f*x + e)^2 + 1)*sinh(f*x + e
)^2 + 2*cosh(f*x + e)^2 + 4*(cosh(f*x + e)^3 + cosh(f*x + e))*sinh(f*x + e) + 1)) - 4*sqrt(2)*((a - b)*cosh(f*
x + e) + (a - b)*sinh(f*x + e))*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*co
sh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)))/((a^2*b - 2*a*b^2 + b^3)*f*cosh(f*x + e)^4 + 4*(a^2*b - 2*a*b^2
+ b^3)*f*cosh(f*x + e)*sinh(f*x + e)^3 + (a^2*b - 2*a*b^2 + b^3)*f*sinh(f*x + e)^4 + 2*(2*a^3 - 5*a^2*b + 4*a
*b^2 - b^3)*f*cosh(f*x + e)^2 + 2*(3*(a^2*b - 2*a*b^2 + b^3)*f*cosh(f*x + e)^2 + (2*a^3 - 5*a^2*b + 4*a*b^2 -
b^3)*f)*sinh(f*x + e)^2 + (a^2*b - 2*a*b^2 + b^3)*f + 4*((a^2*b - 2*a*b^2 + b^3)*f*cosh(f*x + e)^3 + (2*a^3 -
5*a^2*b + 4*a*b^2 - b^3)*f*cosh(f*x + e))*sinh(f*x + e)), -((b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x +
e)^3 + b*sinh(f*x + e)^4 + 2*(2*a - b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 2*a - b)*sinh(f*x + e)^2 + 4
*(b*cosh(f*x + e)^3 + (2*a - b)*cosh(f*x + e))*sinh(f*x + e) + b)*sqrt(-a + b)*arctan(-1/2*sqrt(2)*sqrt(-a + b
)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + si
nh(f*x + e)^2))/((a - b)*cosh(f*x + e) + (a - b)*sinh(f*x + e))) - 2*sqrt(2)*((a - b)*cosh(f*x + e) + (a - b)*
sinh(f*x + e))*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(
f*x + e) + sinh(f*x + e)^2)))/((a^2*b - 2*a*b^2 + b^3)*f*cosh(f*x + e)^4 + 4*(a^2*b - 2*a*b^2 + b^3)*f*cosh(f*
x + e)*sinh(f*x + e)^3 + (a^2*b - 2*a*b^2 + b^3)*f*sinh(f*x + e)^4 + 2*(2*a^3 - 5*a^2*b + 4*a*b^2 - b^3)*f*cos
h(f*x + e)^2 + 2*(3*(a^2*b - 2*a*b^2 + b^3)*f*cosh(f*x + e)^2 + (2*a^3 - 5*a^2*b + 4*a*b^2 - b^3)*f)*sinh(f*x
+ e)^2 + (a^2*b - 2*a*b^2 + b^3)*f + 4*((a^2*b - 2*a*b^2 + b^3)*f*cosh(f*x + e)^3 + (2*a^3 - 5*a^2*b + 4*a*b^2
- b^3)*f*cosh(f*x + e))*sinh(f*x + e))]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Evaluation time:
0.78Error: Bad Argument Type

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maple [C]  time = 0.23, size = 93, normalized size = 1.35 \[ \frac {\mathit {`\,int/indef0`\,}\left (-\frac {\sinh \left (f x +e \right ) \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}}{-b^{2} \left (\sinh ^{6}\left (f x +e \right )\right )+\left (-2 a b -b^{2}\right ) \left (\sinh ^{4}\left (f x +e \right )\right )+\left (-a^{2}-2 a b \right ) \left (\sinh ^{2}\left (f x +e \right )\right )-a^{2}}, \sinh \left (f x +e \right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x)

[Out]

`int/indef0`(-sinh(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/2)/(-b^2*sinh(f*x+e)^6+(-2*a*b-b^2)*sinh(f*x+e)^4+(-a^2-2*a*b
)*sinh(f*x+e)^2-a^2),sinh(f*x+e))/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh \left (f x + e\right )}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(tanh(f*x + e)/(b*sinh(f*x + e)^2 + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {tanh}\left (e+f\,x\right )}{{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(e + f*x)/(a + b*sinh(e + f*x)^2)^(3/2),x)

[Out]

int(tanh(e + f*x)/(a + b*sinh(e + f*x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh {\left (e + f x \right )}}{\left (a + b \sinh ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)/(a+b*sinh(f*x+e)**2)**(3/2),x)

[Out]

Integral(tanh(e + f*x)/(a + b*sinh(e + f*x)**2)**(3/2), x)

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